Thermal physics for IB Physics HL

Reading cue. The thermometer bulb sits in the liquid and not against the coil, and 50.0 W is labelled on the supply, so it is the electrical input rather than the power reaching the liquid.

1. aState State what the 50.0 W on the power supply measures, and state one feature of the drawing that means less than 50.0 W warms the liquid.

   recall2 marks

   Check answer · 2 marks

   1. the rate at which electrical energy is supplied to the heating element
   2. the container is open at the top, so energy escapes there by evaporation and convection (or: the leads, the heater and the container itself absorb energy)

2. bDetermine The liquid has specific heat capacity 2.4 × 10³ J kg⁻¹ K⁻¹. Determine how long the heater must run to raise the temperature of the liquid shown by 1.00 K, assuming all the electrical energy reaches it.

   routine3 marks

   Check answer · 3 marks

   1. E = mcΔT = 0.500 × 2.4 × 10³ × 1.00 = 1.20 × 10³ J
   2. t = E / P = 1.20 × 10³ / 50.0
   3. t = 24.0 s

3. cDetermine The open top is the one surface the hatched insulation does not cover, and 0.42 g of liquid evaporates from it each minute. The specific latent heat of vaporisation is 8.5 × 10⁵ J kg⁻¹. Determine the power this carries away and the resulting error in c.

   demanding4 marks

   Check answer · 4 marks

   1. mass evaporating per second = 4.2 × 10⁻⁴ / 60 = 7.0 × 10⁻⁶ kg s⁻¹
   2. power carried away = 7.0 × 10⁻⁶ × 8.5 × 10⁵ = 6.0 W
   3. only 50.0 − 6.0 = 44.0 W actually warms the liquid, so a value calculated from 50.0 W is too large by a factor 50.0/44.0 = 1.14
   4. c comes out about 14% too high

4. dSuggest The student switches the supply off and keeps reading the thermometer as the liquid cools. Suggest how the cooling readings can be used to correct her value of c, and outline the assumption the correction rests on.

   top of the paper4 marks

   Check answer · 4 marks

   1. after switch-off nothing but the losses is acting, so the rate of fall measured at a given temperature is a direct measure of the loss at that temperature
   2. at that same temperature the heating run gives 50.0 = mc × (rate of rise) + mc × (rate of fall on cooling), so adding the two gradients yields c without needing the loss in watts first
   3. the corrected value of c is smaller than the uncorrected one, because part of the 50.0 W was never warming the liquid
   4. assumes the rate of loss depends only on the excess temperature over the surroundings, so it is the same whether the heater is on or off

Reading cue. The arrows differ in length as well as direction, so the drawing shows a spread of speeds; the rigid walls fix V, which is why warming the gas raises p rather than expanding it.

1. aState State the two quantities shown by or implied by the drawing that cannot change when the container is warmed, and give a reason for each.

   recall2 marks

   Check answer · 2 marks

   1. the volume, 5.0 × 10⁻³ m³, because the container is rigid
   2. the amount of gas, because the container is sealed

2. bDetermine Determine the temperature to which the gas must be warmed for the pressure to reach 3.0 × 10⁵ Pa.

   routine3 marks

   Check answer · 3 marks

   1. V and n are fixed, so p/T is constant: p₁/T₁ = p₂/T₂
   2. T₂ = 290 × (3.0 × 10⁵ / 2.4 × 10⁵) = 362.5 K
   3. T₂ = 363 K, that is about 89 °C

3. cDetermine The arrows represent a spread of molecular speeds. Determine the root mean square speed of a nitrogen molecule at the temperature shown. The molar mass of N₂ is 28.0 g mol⁻¹.

   demanding4 marks

   Show a hint

   The mass of one molecule is the molar mass divided by the Avogadro constant, not the molar mass itself.

   Check answer · 4 marks

   1. average kinetic energy = (3/2)k_BT = 1.5 × 1.38 × 10⁻²³ × 290 = 6.0 × 10⁻²¹ J
   2. mass of one molecule = 28.0 × 10⁻³ / 6.02 × 10²³ = 4.65 × 10⁻²⁶ kg
   3. ½mv²_rms = 6.0 × 10⁻²¹, so v²_rms = 2.58 × 10⁵ m² s⁻²
   4. v_rms = 5.1 × 10² m s⁻¹

4. dExplain A valve is opened briefly and exactly half the nitrogen escapes, the temperature being held at 290 K. Explain how the pressure, the average kinetic energy of a molecule, and the rate of collisions on a given wall each change.

   top of the paper4 marks

   Check answer · 4 marks

   1. p = nRT/V with n halved and V and T unchanged, so the pressure halves to 1.2 × 10⁵ Pa
   2. the average kinetic energy of a molecule is unchanged, because it depends only on the temperature
   3. each remaining molecule moves just as fast and hits just as hard, but there are half as many, so the rate of collisions on a given wall halves
   4. the pressure falls because the collisions are fewer, not because they are gentler